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If work done in stretching a wire by $$1\ mm$$ is $$2\ J$$, the work necessary for stretching another wire of same material, but with double the radius and half the length by $$1\ mm$$ in joule is - 


A
1/4
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B
4
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C
8
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D
16
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Solution

The correct option is D $$16$$
$$ \begin{array}{l} \text { Work done }=U=\frac{1}{2} \times \text { stress } x \text { strain } x \text { velume } \\ \qquad \begin{aligned} U &=\frac{1}{2} x Y \times(\text { strain })^{2} \times \text { volume } \\ U &=\frac{1}{2} \times Y \times \frac{(\Delta l)^{2}}{l^{2}} \times A l \end{aligned} \end{array} $$
$$ \begin{array}{l} =\frac{1}{2} x Y x(\Delta l)^{2} \times \frac{A}{l}=\frac{1}{2} x y x(\Delta \rho)^{2} x \pi \frac{R^{2}}{l} \\ U=\left(\frac{1}{2} x y \pi x(\Delta l)^{2} x\right) \times \frac{R^{2}}{l} \end{array} $$
$$ \begin{array}{l} 2 J=\left[\frac{1}{2} \times y \pi x\left(10^{-3}\right)^{2}\right] \times \frac{R^{2}}{l} \\ U=\left[\frac{1}{2} \times y \pi x\left(10^{-3}\right)^{2}\right] \times \frac{(2 R)^{2}}{\left(\frac{2}{2}\right)} \\ \text { from equation (1) and (2) } \\ U=16 J \end{array} $$

Physics

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